Part 1: A bulb an either glow or burn out so it is problem of binomial distribution. Let X denotes number of bulbs burning out in the first 100 hours of service.

P = Probability of burning out in the first $100$ hours of service $= 0.3$

$$q = 1 – p = 1 – 0.3 = 0.7$$

For binomial distribution, $P (y =x) = \space^n(x p^xq^x)\\ = \space ^3(x (0.3)^x (0.7)^{3-x})$

P (at least 1 bulb is good) = 1 – P (no bulb is good i.e. all 3 bulb burn out)

$$ = 1 – P (x = 3)\\ = 1 – \space^3(3 \times (0.3)^3 \times (0.7)^{3-3})\\ = 1 -1 \times 0.027 \times 1\\ = 0.973$$

Part 2: Let ‘n’ bulbs be needed on each pole to ensure 99% safety that at least one is good after 100 hours.

P (at least 1 bulb is good) $=99\%$

$1 – p$ (no bulb is good) $= 0.99$

$$1 – 0.99 = p (x = n) $$

$0.01 = \space ^n(n (0.3)^n (0.7)^{n - n})\\ 0.01 = (0.3)^n\\ \log 0.01 = n \log 0.3\\ \log \Big(\dfrac {0.01}{0.3}\Big) = n\\ N = 3.825 ≈ 4$

Hence, 4 will be needed on each pole to ensure $99\%$ safety that at least one is good after 100 hours